Proof of the conditional product copula inequalities
Starting point are the following preliminary remarks. According to (3.11), in case (x, t) ∈ E(r13), (y, t) ∈ E(r23), we have if, and only if
(A2.1)
and similarly, we have if, and only if
(A2.2)
By symmetry it suffices to show the inequalities for Case I. The proof is done through exhaustive analysis of all possible disjoint regions that can contain the coordinates (x, t), respectively (y, t). We assume c± > 0 and identify the cases c± = 0 as limiting cases c± → 0. Items related to Theorem 4.1 (4.2) are distinguished using the abbreviation UPC (LPC) for upper (lower) conditional product copula inequalities unless quantities involving c± are used.
-
(1)
(x, t) ∈ E −(r 13) (y, t) arbitrary
(1.1) x + r13 ≥ 0 (UPC), x + r13 < 0 (LPC)
First of all, we note that (LPC). To show this we use the properties of Lemma A1.2 and (3.11) as follows:
(UPC): (P1) ⇒ t ≤ w(x, r13, − 1) ≤ x + r13 − 1 ⇒ (x, t) ∈ L4(r13)
(LPC): (i) using (P6) of Lemma 3.2: t ≤ w(x, r13, − 1) < 1 + r13 − x ⇒ (x, t) ∉ L1(r13)
-
(ii)
using x + r 13 < 0 t ≥ − 1 > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
Furthermore, we must have y + r23 ≥ 0 in Cases (I.a) and (I.b), which is shown as follows:
(UPC) Case (I.a): if y < − r23 then 0 ≤ c+ ≤ ay − bx < − ar23 − br13 = − c+ ≤ 0 a contradiction
(UPC) Case (I.b): y > (−c+ + bx)/a ≥ − (c+ + br13)/a = − r23
(LPC) Case (I.a): if y < − r23, then 0 ≤ c− ≤ ay + bx < − ar23 − br13 = − c− ≤ 0 a contradiction
(LPC) Case (I.b): if y < − r23 then − c− < ay + bx < − ar23 − br13 = − c− a contradiction
If one assumes that t ≤ w(y, r23, − 1), then in virtue of (3.11) because (P1) ⇒ t ≤ w(y, r23, − 1) ≤ y + r23 − 1 ⇒ (y, t) ∈ L4(r23)
We distinguish between three main cases:
-
(a)
c ± t ≥ ay ∓ bx (Case (I.c) and possibly Case (I.b))
Since we have necessarily (UPC), and (LPC), as in the desired inequalities.
-
(b)
c ± t ≤ ay ∓ bx < c ± ⋅ w(y, r 23, − 1) (possible in Cases (I.a) and (I.b))
By the above we have (UPC) and (LPC), which is consistent with the desired inequalities.
-
(c)
c ± t ≤ ay ∓ bx and c ± ⋅ w(y, r 23, − 1) ≤ ay ∓ bx (possible in Cases (I.a) and (I.b))
We derive the inequality (C1.1) t ≤ w(x, r13, − 1) ≤ w(y, r23, − 1) which again implies that (UPC) and (LPC). We use the transformation of variable z = z(y) = ± (ay − c± ⋅ w(y, r23, − 1))/b and its inverse in (A1.3), where the second “±” signs hold for both (UPC) and (LPC). We show (C1.1), that is Δ(x, y) := w(y, r23, − 1) − w(x, r13, − 1) ≥ 0 under the constraints − r13 ≤ x ≤ z(y) (UPC), z(y) ≤ x < − r13 (LPC), − r23 ≤ y ≤ 1. By (P3) of Lemma A1.2 the function w(x, r13, − 1) is increasing in x (UPC), decreasing in x (LPC). It follows that Δ(x, y) ≥ Δ*(y, z(y)) := w(y, r23, − 1) − w(z(y), r13, − 1). Using the transformation z(y) we can write w(y, r23, − 1) = (ay ∓ bz(y))/c±. Making also use of the inverse y = y(z) we see that (use the first identity in (A1.1))
which implies that Δ(x, y) ≥ Δ*(y, z(y)) = Δ*(z) ≥ 0 as desired.
(1.2) x + r13 < 0 (UPC), x + r13 ≥ 0 (LPC)
We have (UPC), (LPC), which is shown as follows:
(UPC): (i) using (P6) of Lemma A1.2: t ≤ w(x, r13, − 1) < 1 + r13 − x ⇒ (x, t) ∉ L1(r13)
-
(ii)
using x + r 13 < 0: t ≥ − 1 > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
(LPC): (P1) of Lemma A1.2 ⇒ t ≤ w(x, r13, − 1) ≤ x + r13 − 1 ⇒ (x, t) ∈ L4(r13)
We distinguish between two main cases:
-
(a)
c ± t ≤ ay ∓ bx (Case (I.a) and possibly Case (I.b))
We have (UPC), (LPC) as desired
-
(b)
ay ∓ bx ≤ c ± t ≤ c ± ⋅ w(x, r 13, − 1) (Case (I.c) and possibly Case (I.b))
We show , which is consistent with the desired results (UPC) and (LPC). It suffices to show the following inequalities:
Case 1: x ≥ − (b + c+r13)/a (UPC), x ≤ (b − c−r13)/a (LPC)
(C1.2.1) t ≤ w(x, r13, − 1) ≤ w(y, r23, − 1), and y + r23 ≤ 0
Case 2: x ≤ − (b + c+r13)/a (UPC), x ≥ (b − c−r13)/a (LPC)
(C1.2.2) c± ⋅ w(y, r23, 1) ≤ ay ∓ bx ≤ c±t ≤ c± ⋅ w(x, r13, − 1), and y − r23 ≤ 0
Indeed, if these conditions hold, then in virtue of (3.11) as follows:
Case 1: (i) using (P6) of Lemma A1.2: t ≤ w(y, r23, − 1) < 1 + r23 − y ⇒ (y, t) ∉ L1(r23)
-
(ii)
using y + r 23 ≤ 0: t ≥ − 1 ≥ y + r 23 − 1 ⇒ (y, t) ∉ L 4(r 23)
Case 2: (i) using y − r23 ≤ 0: t ≤ 1 ≤ 1 + r23 − y ⇒ (y, t) ∉ L1(r23)
-
(ii)
using (P5) of Lemma A1.2: t ≥ w(y, r 23, 1) > y + r 23 − 1 ⇒ (y, t) ∉ L 4(r 23)
To show the above inequalities we use the transformation z = z(x) = (c± ⋅ w(x, r13, − 1) ± bx)/a.
Case 1: The function z(x) is increasing in the interval − (b + c±r13)/a ≤ x < − r13 (UPC), respectively z(x) is decreasing in the interval − r13 ≤ x ≤ (b − c−r13)/a (LPC):
In particular, by the restriction defining case (b), we have necessarily y ≤ z(x) ≤ z(−r13) = (−c± ∓ br13)/a = − r23 ⇒ y + r23 ≤ 0.
Now, we show that Δ(x, y) = w(y, r23, − 1) − w(x, r13, − 1) ≥ 0 under the constraints, − 1 ≤ y ≤ z(x), − (b + c+r13)/a ≤ x < − r13 (UPC), − r13 ≤ x ≤ (b − c−r13)/a (LPC). Since w(y, r23, − 1) is decreasing in y by (P3) of Lemma A1.2, we have Δ(x, y) ≥ Δ*(x, z(x)) := w(z(x), r23, − 1) − w(x, r13, − 1) = w(z(x), r23, − 1) − (az(x) ∓ bx)/c±.
Since z(x) is increasing and z(−r13) = − r23, z(−(b + c+r13)/a) = − 1 (UPC), respectively z(x) is decreasing and z(−r13) = − r23, z((b − c−r13)/a) = − 1, its inverse in (A1.3) takes the “+” sign (UPC) respectively the “-“ sign (LPC), that is . Insert it into the function Δ*(z) = Δ*(x(z), z) = w(z, r23, − 1) − (az ∓ bx(z))/c± to see that Δ(x, y) ≥ Δ*(x, z(x)) = Δ*(z) = 0 as desired.
Case 2: Similarly to Case 1, one shows that z(x) is decreasing in the interval − 1 ≤ x ≤ − (b + c+r13)/a (UPC), respectively z(x) is increasing in the interval (b − c−r13)/a ≤ x ≤ 1 (LPC), and in particular y ≤ z(x) ≤ z(−1) = − (±c±r13 + b)/a ≤ r23, where the last inequality is true because ar23 + b + c+r13 = (b + c+) ⋅ (1 + r13) ≥ 0 (UPC), ar23 + b − c−r13 ≥ c− ⋅ (1 − r13) ≥ 0 (LPC).
Now, we show that Δ(x, y) = (ay ∓ bx)/c± − w(y, r23, 1) ≥ 0 under the constraints − 1 ≤ y ≤ z(x), − 1 ≤ x ≤ − (b + c+r13)/a (UPC), (b − c−r13)/a ≤ x ≤ 1 (LPC). The function h(y) = ± (ay/c± − w(y, r23, 1)) is decreasing (UPC), respectively increasing (LPC), in the interval − 1 ≤ y ≤ − (b ± c±r13)/a:
It follows that Δ(x, y) ≥ Δ*(x, z(x)) = (az(x) ∓ bx)/c± − w(z(x), r23, 1). Since z(x) is decreasing (UPC), respectively increasing (LPC), and z(∓(b ± c±r13)/a) = − 1, z(∓1) = ∓ (b ± c±r13)/a, its inverse in (A1.3) takes the “-“ sign (UPC) respectively the “+“ sign (LPC), that is we have . Insert this expression into the function Δ*(z) = Δ*(x(z), z) = (az ∓ bx(z))/c± − w(z, r23, 1) to see that Δ(x, y) ≥ Δ*(x, z(x)) = Δ*(z) = 0.
-
(2)
(y, t) ∈ E −(r 23) (x, t) arbitrary
(2.1) y + r23 < 0
We have , which follows from Lemma A1.2 and (3.11) as follows:
-
(i)
using (P6) of Lemma A1.2: t ≤ w(y, r 23, − 1) < 1 + r 23 − y ⇒ (y, t) ∉ L 1(r 23)
-
(ii)
using y + r 23 < 0: t ≥ − 1 > y + r 23 − 1 ⇒ (y, t) ∉ L 4(r 23)
Furthermore, we must have x + r13 < 0 (UPC), respectively x + r13 ≥ 0 (LPC), in Cases (I.a) and (I.b), which is shown as follows:
(UPC) Case (I.a): ay − bx ≥ c+ ⇒ x ≤ (ay − c+)/b < − (c+ + ar13)/b ≤ − r13, where the last inequality follows from ar23 − br13 + c+ = 2c+ ≥ 0
(UPC) Case (I.b): ay − bx > − c+ ⇒ x < (c+ + ay)/b < (c+ − ar23)/b = − r13
(LPC) Case (I.a): if x < − r13, then 0 ≤ c− ≤ ay + bx < − ar23 − br13 = − c− ≤ 0, a contradiction
(LPC) Case (I.b): if x < − r13, then − c− < ay + bx < − ar23 − br13 = − c−, a contradiction
In these Cases, if t ≤ w(x, r13, − 1), then (UPC), (LPC):
(UPC): (i) using (P6) of Lemma 3.2: t ≤ w(x, r13, − 1) < 1 + r13 − x ⇒ (x, t) ∉ L1(r13)
-
(ii)
using x + r 13 < 0: t ≥ − 1 > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
(LPC): (P1) of Lemma A1.2 ⇒ x + r13 ≥ 0 ⇒ w(x, r13, − 1) ≤ x + r13 − 1 ⇒ (x, t) ∈ L4(r13)
We have three main cases:
-
(a)
c ± t ≥ ay ∓ bx (Case (I.c) and possibly Case (I.b))
We have necessarily (UPC), (LPC)
-
(b)
c ± t ≤ ay ∓ bx < c ± ⋅ w(x, r 13, − 1) (possible in Cases (I.a) and (I.b))
By the above we have (UPC), (LPC), which is consistent with the desired inequalities.
-
(c)
c ± t ≤ ay ∓ bx and c ± ⋅ w(y, r 23, − 1) ≤ ay ∓ bx (possible in Cases (I.a) and (I.b))
We derive the inequality (C2.1) t ≤ w(y, r23, − 1) ≤ w(x, r13, − 1), which again implies (UPC), (LPC). We use the transformation of variable z = z(x) = (c± ⋅ w(x, r13, − 1) ± bx)/a and its inverse in (A1.3), where the second “±” signs hold for both (UPC) and (LPC). We show that Δ(x, y) := w(x, r13, − 1) − w(y, r23, − 1) ≥ 0 under the constraints z(x) ≤ y < − r23 , − 1 ≤ x < − r13 (UPC), − r13 ≤ x ≤ 1 (LPC). By (P3) of Lemma A1.2 the function w(y, r23, − 1) is decreasing in y. It follows that Δ(x, y) ≥ Δ*(x, z(x)) = w(x, r13, − 1) − w(z(x), r23, − 1) = (az(x) ∓ bx)/c± − w(z(x), r23, − 1).
Now, insert the inverse x = x(z) into Δ*(z) = Δ*(x(z), z) = (az ∓ bx(z))/c± − w(z, r23, − 1) to see that as desired.
(2.2) y + r23 ≥ 0
Property (P1) implies t ≤ w(y, r23, − 1) ≤ y + r23 − 1 ⇒ (y, t) ∈ L4(r23) ⇒
We distinguish between two main cases:
-
(a)
c ± t ≤ ay ∓ bx (Case (I.a) and possibly Case (I.b))
We have necessarily (UPC), (LPC).
-
(b)
ay ∓ bx ≤ c ± t ≤ c ± ⋅ w(x, r 13, − 1) (Case (I.c) and possibly Case (I.b))
We show that (UPC), (LPC), which is consistent with (UPC), (LPC). It suffices to show the following inequalities:
Case 1: y ≤ (a − c±r23)/b
(C2.2.1) t ≤ w(y, r23, − 1) ≤ w(x, r13, − 1, and x + r13 ≥ 0 (UPC), x + r13 ≤ 0 (LPC)
Case 2: y ≥ (a − c±r23)/b
(C2.2.2) w(x, r13, 1) ≤ (ay − bx)/c ≤ t ≤ w(y, r23, − 1), and x − r13 ≥ 0 (UPC), x − r13 ≤ 0 (LPC)
Indeed, if these conditions hold, then (UPC), (LPC) as follows:
(UPC) Case 1: Property (P1) implies t ≤ w(x, r13, − 1) ≤ x + r13 − 1 ⇒ (x, t) ∈ L4(r13)
(LPC) Case 1:
-
(i)
using (P6) of Lemma 3.2: t ≤ w(x, r 13, − 1) < 1 + r 13 − x ⇒ (x, t) ∉ L 1(r 13)
-
(i)
using x + r 13 ≤ 0: t ≥ − 1 ≥ x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
(UPC) Case 2: Property (P2) implies t ≥ w(x, r13, 1) ≥ 1 + r13 − x ⇒ (x, t) ∈ L1(r13)
(LPC) Case 2:
-
(i)
using x − r 13 ≤ 0: t ≤ 1 ≤ 1 + r 13 − x ⇒ (x, t) ∉ L 1(r 13)
-
(i)
using (P5) of Lemma 3.2: t ≥ w(x, r 13, 1) > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
To show the above inequalities we use the transformation z = z(y) = ± (ay − c± ⋅ w(y, r23, − 1))/b.
Case 1: The function z(y) is increasing (UPC) (respectively decreasing (LPC)) in the interval − r23 ≤ y ≤ (a − c±r23)/b: , hence x ≥ z(y) ≥ z(−r23) = (c+ − ar23)/b = − r13 (UPC), x ≤ z(y) ≤ z(−r23) = (ar23 − c−)/b = − r13 (LPC). Now, we show that Δ(x, y) = w(x, r13, − 1) − w(y, r23, − 1) ≥ 0 under the constraints z(y) ≤ x ≤ 1 (UPC), − 1 ≤ x ≤ z(y) (LPC), − r23 ≤ y ≤ (a − c±r23)/b.
Since w(x, r13, − 1) is increasing in x (UPC) (decreasing in x (LPC)) by (P3) of Lemma A1.2, we have Δ(x, y) ≥ Δ*(y, z(y)) = w(z(y), r13, − 1) − w(y, r23, − 1) = w(z(y), r13, − 1) − (ay ∓ bz(y))/c.
Since z(y) is increasing and z(−r23) = − r13, z((a − c+r23)/b) = 1 (UPC), respectively z(y) is decreasing and z(−r23) = − r13, z((a − c−r23)/b) = − 1 (LPC), its inverse takes the “-“ sign for both (UPC) and (LPC), that is . Insert it into the function Δ*(z) = Δ*(y(z), z) = w(z, r23, − 1) − (ay(z) ∓ bz)/c± to see that Δ(x, y) ≥ Δ*(y, z(y)) = Δ*(z) = 0 as desired.
Case 2: Similarly to Case 1, one shows that z(y) is decreasing (UPC) (respectively increasing (LPC)) in the interval (a − c±r23)/b ≤ y ≤ 1, and in particular x ≥ z(y) ≤ z(1) = (a − c+r23)/b ≥ r13 (UPC), x ≤ z(y) ≤ z(1) = (c−r23 − a)/b ≤ r13 (LPC), where the last inequalities are true because a − br13 − c+r23 ≥ c+ ⋅ (1 − r23) ≥ 0 (UPC), br13 + a − c−r23 ≥ c− ⋅ (1 − r23) ≥ 0 (LPC). Now, we show that Δ(x, y) = (ay ∓ bx)/c± − w(x, r13, 1) ≥ 0 under the constraints z(y) ≤ x ≤ 1 (UPC), − 1 ≤ x ≤ z(y) (LPC), (a − c±r23)/b ≤ y ≤ 1. The function h(x) = w(x, r13, 1) ± bx/c is decreasing in the interval (a − c+r23)/b ≤ x ≤ 1 (UPC), respectively increasing in the interval − 1 ≤ x ≤ (c−r23 − a)/b (LPC):
It follows that Δ(x, y) ≥ Δ*(y, z(y)) = (ay ∓ bz(y))/c± − w(z(y), r13, 1). Since z(y) is decreasing and z((a − c+r23)/b) = 1, z(1) = (a − c+r23)/b (UPC), respectively z(y) is increasing and z((a − c−r23)/b) = − 1, z(1) = (c−r23 − a)/b (LPC), its inverse takes the “+” sign for both (UPC) and (LPC), that is . Insert it into Δ*(z) = Δ*(y(z), z) = (ay(z) ∓ bz)/c± − w(z, r13, 1) to see that Δ(x, y) ≥ Δ*(y, z(y)) = Δ*(z) = 0.
-
(3)
(x, t) ∈ E(r 13) and (y, t) ∈ E(r 23)
The validity of the inequality in the Cases (I.a) and (I.b) follows from (A2.1). In Case (I.c) we have (ay − bx)/c+ ≤ − 1 ≤ t, hence . Similarly, the validity of the inequality in the Cases (I.a) and (I.b) follows from (A2.2). In Case (I.c) we have (ay + bx)/c− ≤ − 1 ≤ t, hence .
-
(4)
(x, t) ∈ E +(r 13), (y, t) arbitrary
(4.1) x − r13 < 0 (UPC), x − r13 ≥ 0 (LPC)
We first note that (UPC), (LPC):
(UPC): (i) using x − r13 < 0: t ≤ 1 < 1 + r13 − x ⇒ (x, t) ∉ L1(r13)
-
(ii)
using (P5) of Lemma A1.2: t ≥ w(x, r 13, 1) > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
(LPC): (P2) of Lemma A1.2: t ≥ w(x, r13, 1) ≥ 1 + r13 − x ⇒ (x, t) ∈ L1(r13) ⇒
We must have y − r23 < 0 in Cases (I.b) and (I.c). We argue as follows:
(UPC) Case (I.b): x < r13 ⇒ y < (c + bx)/a ≤ (c + br13)/a = r23
(UPC) Case (I.c): if y ≥ r23, − x > − r13 then ay − bx ≥ ar23 − br13 = c in contradiction to the assumption ay − bx ≤ − c defining Case (I.c).
(LPC) Case (I.b): − x ≤ − r13 ⇒ y < (c− − bx)/a ≤ (c− − br13)/a = r23
(LPC) Case (I.c): if y ≥ r23, x ≥ r13 then ay + bx ≥ ar23 + br13 = c− in contradiction to the assumption ay + bx ≤ − c− defining Case (I.c).
In this situation, if one assumes t ≥ w(y, r23, 1), then in virtue of (3.11):
(UPC): (i) using y − r23 < 0: t ≤ 1 < 1 + r23 − y ⇒ (y, t) ∉ L1(r23)
-
(ii)
using (P5) of Lemma 3.2: t ≥ w(y, r 23, 1) > y + r 23 − 1 ⇒ (y, t) ∉ L 4(r 23)
(LPC): (i) using y − r23 < 0: t ≤ 1 < 1 + r23 − y ⇒ (y, t) ∉ L1(r23)
-
(ii)
using (P5) of Lemma 3.2: t ≥ w(y, r 23, 1) > y + r 23 − 1 ⇒ (y, t) ∉ L 4(r 23)
We have three main cases:
-
(a)
c ± t ≤ ay ∓ bx (Case (I.a) and possibly Case (I.b))
We have necessarily (UPC), (LPC).
-
(b)
c ± t ≥ ay ∓ bx > c ± ⋅ w(y, r 23, 1) (possible in Cases (I.b) and (I.c))
By the above we have (UPC), (LPC), which is consistent with the desired inequalities
-
(c)
c ± t ≥ ay ∓ bx and ay ∓ bx ≤ c ± ⋅ w(y, r 23, 1) (possible in Cases (I.b) and (I.c))
We derive the inequality (C4.1) t ≥ w(x, r13, 1) ≥ w(y, r23, 1), which again implies that (UPC), (LPC). We use the transformation z = z(y) = ± (ay − c± ⋅ w(y, r23, 1))/b and its inverse , where the second “±” signs hold for both (UPC) and (LPC). We show that Δ(x, y) = w(x, r13, 1) − w(y, r23, 1) ≥ 0 under the constraints z(y) ≤ x < r13 (UPC), r13 ≤ x ≤ z(y) (LPC), − 1 ≤ y < r23. By (P4) of Lemma A1.2 the function w(x, r13, 1) is increasing in x (UPC), respectively decreasing in x (LPC). It follows that Δ(x, y) ≥ Δ*(y, z(y)) = w(z(y), r13, 1) − w(y, r23, 1) = w(z(y), r13, 1) − (ay ∓ bz(y))/c±.
Insert the inverse y = y(z) into Δ*(z) = Δ*(y(z), z) = w(z, r13, 1) − (ay(z) ∓ bz)/c± to see that as desired.
(4.2) x − r13 ≥ 0 (UPC), x − r13 < 0 (LPC)
We first note that (UPC), (LPC):
(UPC): Property (P2) implies t ≥ w(x, r13, 1) ≥ 1 + r13 − x ⇒ (x, t) ∈ L1(r13)
(LPC): (i) using x − r13 < 0: t ≤ 1 < 1 + r13 − x ⇒ (x, t) ∉ L1(r13)
-
(ii)
using (P5) of Lemma 3.2: t ≥ w(x, r 13, 1) > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
We distinguish between two main cases:
-
(a)
c ± t ≥ ay ∓ bx (Case (I.c) and possibly Case (I.b))
We have necessarily (UPC), (LPC)
-
(b)
ay ∓ bx ≥ c ± t ≥ c ± ⋅ w(x, r 13, 1) (Case (I.a) and possibly Case (I.b))
We show that , which is consistent with the desired inequalities.
We show Case 1: x ≤ (b + c+r13)/a (UPC), x ≥ (c−r13 − b)/a (LPC) (C4.2.1) t ≥ w(x, r13, 1) ≥ w(y, r23, 1), and y − r23 ≥ 0 (UPC), y − r23 > 0 (LPC)
Case 2: x ≥ (b + c+r13)/a (UPC), x ≤ (c−r13 − b)/a (LPC) (C4.2.2) c± ⋅ w(x, r13, 1) ≤ c±t ≤ ay ∓ bx < c± ⋅ w(y, r23, − 1), and y + r23 ≥ 0
Indeed, if these conditions hold, then we have in virtue of (3.11) as follows:
Case 1: (P2) ⇒ t ≥ w(y, r23, 1) ≥ 1 + r23 − y ⇒ (y, t) ∈ L1(r23)
Case 2: (P1) ⇒ t < w(y, r23, − 1) ≤ y + r23 − 1 ⇒ (y, t) ∈ L4(r23)
To show the above inequalities we use the transformation z = z(x) = (c± ⋅ w(x, r13, 1) ± bx)/a.
Case 1: The function z(x) is increasing in the interval r13 ≤ x ≤ (b + c+r13)/a (UPC), respectively decreasing in the interval (c−r13 − b)/a ≤ x < r13 (UPC): .
In particular y ≥ z(x) > z(r13) = (c+ + br13)/a = r23 (UPC), y ≥ z(x) > z(r13) = (c− − br13)/a = r23 (LPC). Now, we show that Δ(x, y) = w(x, r13, 1) − w(y, r23, 1) ≥ 0 under the constraints z(x) ≤ y ≤ 1, r13 ≤ x ≤ (b + c+r13)/a (UPC), (c−r13 − b)/a ≤ x < r13 (LPC). Since w(y, r23, 1) is decreasing in y by (P4) of Lemma 3.2, we have Δ(x, y) ≥ Δ*(x, z(x)) = w(x, r13, 1) − w(z(x), r23, 1) = (az(x) ∓ bx)/c± − w(z(x), r23, 1).
Since z(x) is increasing and z(r13) = r23, z((b + c+r13)/a) = 1 (UPC), respectively z(x) is decreasing, z(r13) = r23, z((c−r13 − b)/a) = 1 its inverse in (A1.3) takes the “-“ sign (UPC), respectively the “+” sign (LPC), that is . Insert it into Δ*(z) = Δ*(x(z), z) = (az ∓ bx(z))/c± − w(z, r23, 1) to see that Δ(x, y) ≥ Δ*(x, z(x)) = Δ*(z) = 0 as desired.
Case 2: Similarly to Case 1, one shows that z(x) is decreasing in the interval (b + c+r13)/a ≤ x ≤ 1 (UPC), respectively z(x) is increasing in the interval − 1 ≤ x ≤ (c−r13 − b)/a (LPC). In particular y ≥ z(x) ≥ z(1) = (b + c+r13)/a ≥ − r23 (UPC), y ≥ z(x) ≥ z(−1) = (b − c−r13)/a ≥ − r23 (LPC), where the last inequalities are true because ar23 + b + c+r13 = (b + c+) ⋅ (1 + r13) ≥ 0 (UPC), ar23 + b − c−r13 ≥ c− ⋅ (1 − r13) ≥ 0 (LPC).
Now, we show that Δ(x, y) = w(y, r23, − 1) − (ay ∓ bx)/c± ≥ 0 under the constraints z(x) ≤ y ≤ 1, (b + c+r13)/a ≤ x ≤ 1 (UPC), − 1 ≤ x ≤ (c−r13 − a)/b (LPC). The function h(y) = w(y, r23, − 1) − ay/c± is increasing in the interval (b ± c±r13)/a ≤ y ≤ 1:
It follows that Δ(x, y) ≥ Δ*(x, z(x)) = w(z(x), r23, − 1) − (az(x) ∓ bx)/c±. Since z(x) is decreasing and z((b + c+r13)/a) = 1, z(1) = (b + c+r13)/a (UPC), respectively z(x) is increasing and z((c−r13 − b)/a) = 1, z(−1) = (a − c−r23)/b (LPC), its inverse takes the “+” sign (UPC), respectively the “-“ sign (LPC), that is . Insert it into Δ*(z) = Δ*(x(z), z) = w(z, r23, 1) − (az ∓ bx(z))/c± to see that Δ(x, y) ≥ Δ*(x, z(x)) = Δ*(z) = 0.(5)
(5.1)
Property (P2) implies t ≥ w(y, r23, 1) ≥ 1 + r23 − y ⇒ (y, t) ∈ L1(r23) ⇒
We must have x − r13 ≥ 0 (UPC), x − r13 ≥ 0 (LPC), in Cases (I.b) and (I.c):
(UPC) Case (I.b): x > (ay − c+)/b ≤ (ar23 − c+)/b = r13
(UPC) Case (I.c): if − x > − r13, y ≥ r23 then ay − bx > ar23 − br13 = c+ in contradiction to the assumption ay − bx ≤ − c+ defining Case (I.c).
(LPC) Case (I.b): − y ≤ − r23 ⇒ x < (c− − ay)/b ≤ (c− − ar23)/b = r13
(LPC) Case (I.c): if x ≥ r13, y ≥ r23 then ay + bx ≥ ar23 + br13 = c− in contradiction to the assumption ay + bx ≤ − c− defining Case (I.c).
In this situation, if t ≥ w(x, r13, 1), then (UPC), (LPC):
(UPC): using (P2) of Lemma A1.2: t ≥ w(x, r13, 1) ≥ 1 + r13 − x ⇒ (x, t) ∈ L1(r13)
(LPC):
-
(i)
using x − r 13 < 0: t ≤ 1 < 1 + r 13 − x ⇒ (x, t) ∉ L 1(r 13)
-
(ii)
using (P5) of Lemma A1.2: t ≥ w(x, r 13, 1) > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
We distinguish between three main cases:
-
(a)
c ± t ≤ ay ∓ bx (Case (I.a) and possibly Case (I.b))
We have necessarily (UPC), (LPC)
-
(b)
c ± t ≥ ay ∓ bx > c ± ⋅ w(x, r 13, 1) (possible in Cases (I.b) and (I.c))
By the above we have (UPC), (LPC), which is consistent with the desired inequalities
-
(c)
c ± t ≥ ay ∓ bx and ay ∓ bx ≤ c ± ⋅ w(x, r 13, 1) (possible in Cases (I.b) and (I.c))
We derive the inequality (C5.1) t ≥ w(y, r23, 1) ≥ w(x, r13, 1), which again implies (UPC), (LPC). We use the transformation z = z(x) = (c ⋅ w(x, r13, 1) ± bx)/a and its inverse in (A1.3) to show that Δ(x, y) = w(y, r23, 1) − w(x, r13, 1) ≥ 0 under the constraints r23 ≤ y ≤ z(x), r13 ≤ x ≤ 1 (UPC), − 1 ≤ x < r13 (LPC). By (P4) of Lemma A1.2 the function w(y, r23, 1) is decreasing in y, hence Δ(x, y) ≥ Δ*(x, z(x)) = w(z(x), r23, 1) − w(x, r13, 1) = w(z(x), r23, 1) − (az(x) ÷ bx)/c±.
Insert the inverse x = x(z) into Δ*(z) = Δ*(x(z), z) = w(z, r23, 1) − (az ∓ bx(z))/c± to see that as desired.(5.2)
We first note that :
-
(i)
using y − r 23 < 0: t ≤ 1 < 1 + r 23 − y ⇒ (y, t) ∉ L 1(r 23)
-
(ii)
using (P5) of Lemma 3.2: t ≥ w(y, r 23, 1) > y + r 23 − 1 ⇒ (y, t) ∉ L 4(r 23)
We distinguish between two main cases:
-
(a)
c ± t ≥ ay ∓ bx (Case (I.c) and possibly Case (I.b))
We have necessarily (UPC), (LPC)
-
(b)
c ± ⋅ w(y, r 23, 1) ≤ c ± t ≤ ay ∓ bx (Case (I.a) and possibly Case (I.b))
We show that (UPC), (LPC), which is consistent with the desired inequalities (UPC), (LPC)
Case 1 y ≥ (c±r23 − a)/b
(C5.2.1) t ≥ w(y, r23, 1) ≥ w(x, r13, 1), and x − r13 < 0 (UPC), x − r13 > 0 (LPC)
Case 2 y ≤ (c±r23 − a)/b
(C5.2.2) c± ⋅ w(y, r23, 1) ≤ c±t ≤ ay ∓ bx < c± ⋅ w(x, r13, − 1), x + r13 < 0 (UPC), x + r13 ≥ 0 (LPC)
Indeed, if these conditions hold, then in virtue of (3.11) as follows:
UPC) Case 1 (i) using x − r13 < 0: t ≤ 1 < 1 + r13 − x ⇒ (x, t) ∉ L1(r13)
-
(ii)
using (P5) of Lemma 3.2: t ≥ w(x, r 13, 1) > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
(LPC) Case 1: (P2) ⇒ t ≥ w(x, r13, 1) ≥ 1 + r13 − x ⇒ (x, t) ∈ L1(r13)
(UPC) Case 2: (i) using (P6) of Lemma 3.2: t ≤ w(x, r13, − 1) < 1 + r13 − x ⇒ (x, t) ∉ L1(r13)
-
(ii)
using x + r 13 < 0 t ≥ − 1 > x + r 13 − 1 ⇒ (x, t) ∉ L 4(r 13)
(LPC) Case 2: (P1) ⇒ t < w(x, r13, − 1) ≤ x + r13 − 1 ⇒ (x, t) ∈ L4(r13)
To show the above inequalities we use the transformation z = z(y) = ± (ay − c± ⋅ w(y, r23, 1))/b.
Case 1: Since the function z(y) is increasing (UPC) (respectively decreasing (LPC)) in the interval (c±r23 − a)/b ≤ y < r23, we have in particular x ≤ z(y) < z(r23) = (ar23 − c+)/b = r13 (UPC), x ≥ z(y) > z(r23) = (c− − ar23)/b = r13 (LPC). We show that Δ(x, y) = w(y, r23, 1) − w(x, r13, 1) ≥ 0 under the constraints − 1 ≤ x ≤ z(y) (UPC), z(y) ≤ x ≤ 1 (LPC), (c±r23 − a)/b ≤ y < r23. With (P4) of Lemma A1.2, w(x, r13, 1) is increasing in x (UPC), decreasing in x (LPC), hence Δ(x, y) ≥ Δ*(y, z(y)) = w(y, r23, 1) − w(z(y), r13, 1) = (ay ∓ bz(y))/c± − w(z(y), r13, 1).
Since z(y) is increasing and z(r23) = r13, z((c+r23 − a)/b) = − 1 (UPC), respectively z(y) is decreasing and z(r23) = r13, z((c−r23 − a)/b) = 1 (LPC), its inverse in (A1.3) takes the “+” sign: . Insert into Δ*(z) = Δ*(y(z), z) = (ay(z) ∓ bz)/c± − w(z, r23, 1) to see that Δ(x, y) ≥ Δ*(y, z(y)) = Δ*(z) = 0 as desired.
Case 2: One shows that z(y) is decreasing (UPC) (respectively increasing (LPC)) in the interval − 1 ≤ y ≤ (c±r23 − a)/b. In particular x ≤ z(y) ≤ z(−1) = (c+r23 − a)/b ≤ − r13 (UPC), x ≥ z(y) ≥ z(−1) = (a − c−r23)/b ≥ − r13 (LPC), where the last inequalities are true because a − br13 − c+r23 ≥ c+ ⋅ (1 − r23) ≥ 0 (UPC), a + br13 − c−r23 ≥ c− ⋅ (1 − r23) ≥ 0 (LPC).
Now, we show that Δ(x, y) = w(x, r13, − 1) − (ay ∓ bx)/c± ≥ 0 under the constraints − 1 ≤ x ≤ z(y) (UPC), z(y) ≤ x ≤ 1 (LPC), − 1 ≤ y ≤ (c±r23 − a)/b. Since the function h(x) = w(x, r13, − 1) ± bx/c is decreasing (UPC), respectively increasing (LPC), we have Δ(x, y) ≥ Δ*(y, z(y)) = w(z(y), r13, − 1) − (ay ∓ bz(y))/c±. Since z(y) is decreasing and z((c+r23 − a)/b) = − 1, z(−1) = (c+r23 − a)/b (UPC), respectively z(y) is increasing and z((c−r23 − a)/b) = 1, z(−1) = (a − c−r23)/b (LPC), its inverse in (A1.3) takes the “-“ sign, that is . Insert it into Δ*(z) = Δ*(y(z), z) = w(z, r13, − 1) − (ay(z) − bz)/c to see that Δ(x, y) ≥ Δ*(y, z(y)) = Δ*(z) = 0 as desired.